Fill in the missing values in the ANOVA tables below and answer the questions relatingto each table and the provided plots.Df Sum Sq Mean Sq F-value p-valueGroup227.1597.2e-06Residuals 94189.58TotalsTable 1: Q9) ANOVA table.(a) How many treatment groups are there in this experiment and what was the totalsample size?(b) What are Ho and HA relating to the F statistic in this table?(c) Let a =0.001. Given the output of this table, what decision would you makeabout the hypotheses?(d) What does the p-value in this table represent as a probability with respect to theANOVA hypotheses?(e) A Shapiro-Wilk test returned a p-value of 0.82. Using this information and theplots below, is there evidence that this analysis was not correctly done?Normal Q-QResiduals vs Fitted0130813....00039039081T81°T-2-1110.010.511.011.5Theoretical QuantilesFitted valuesaov(ANOV_X ~ group)aov(ANOV_X ~ group)Figure 1: Q9) QQ-plot.Figure 2: Q9) Residual plot.Standardized residuals-2E--11 2ResidualsO 00 00 MDOD O O O

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The values in the column for the degrees of freedom are based on the number of k population means, or treatments, and the total number of observations, N. In the sum of squares column, SSTT represents a measure of differences among the sample means. The value of SSE is a measure of variability within the k samples. Note that the values in the mean square column are calculated based on the sum of squares and degrees of freedom for the treatments row and error row. These mean square values are then used to calculate the F test statistic. The P-value is based on the F test statistic with k - 1 numerator degrees of freedom and N - k denominator degrees of freedom. The given data are below. Treatment 1 Treatment 2 Treatment 3 Treatment 4 9 5 5 6. 3 1 14 7 4 2 1 4. 3 12 3 4 9 2 2 9 1 2 14 8 3 3 3 3 3 4. 11 11 7 5 10 1 10
You have performed a one-way ANOVA to compare 6 groups with a total of 54 observations. You have have the following ANOVA table - fil in the missing values Source ss df MS F P.value Group (Between) 1000 Number 200 29.63 NOT CALCULATED Error (Within) 324 Number 6.75 Total 1324 Number
we look at a random sample of 1000 united flights in the month of December comparing the actual arrival time to the scheduled arrive time. computer output of the descriptive statistics for the difference in actual and expected arrival time of these 1000 flights are shown below : N=1000 Mean =4.06 St Dev. = 45.4 SE Mean =1.44 Min= -35 Q1 = -14 Median = -5 Q3 =9 Max =871 what is the sample mean difference in actual and expected arrival times? what is the standard deviation of the difference ? x bar = and s= ?
What are the values for SS and variance for the following sample of n = 3 scores? Sample: 8, 5, 2
what is the value in cell D? Round to 2 decimal places. How many total observations were there in the study? Hint: use df(total) to figure out the number of total observations. The following computer output is for an analysis of variance in which yields (bu/acre) of different varieties of oats were compared. Answer the following questions based on this ANOVA table. Source df SS MS F ratio Prob Between groups 2 76.90 38.45 D 0.63 Within groups 11 859.81 78.16 Total 13 936.17
Consider the following ANOVA table below. Some parts are blank intentionally. The number of observations is 249. Source Model (Regression) Residual (Error) Total The MSE is: SS 1,332 DF MS 921 Round your answer to two decimal places.
How is the sample size related to the margin of error and confidence level? What isthe impact of S.D. on sample size? Comments with justification. Also fill in theblanks showing necessary calculations in each case: Margin of error S .D. Confidence Level Sample Size 0.4274 5.33 95% ________ 0.8548. 4.79 ___________. 209 0.7948 ______ 90% 249 0.2129 5.79 __________. 2841 __________. 2.54 90% 3720
A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the following table. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants. Sample mean Sample variance Sample standard deviation Plant 1 Atlanta 84 76 81 76 70 81 78 25.2 5.02 Plant 2 Plant 3 Dallas Seattle 72 74 72 74 70 88 75 42.8 6.54 60 64 62 68 76 66 66 32.0 5.66
Find the percentage of the total area under the standard normal curve between the following z-scores. z=1.44 and z=2.83 Click here for page 1 of the Areas under the Normal Curve Table Click here for page 2 of the Areas under the Normal Curve Table The percentage of the total area between z=1 44 and z=2.83 is% (Round to two decimal places as needed.)
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